# Card sharp

PBE6
Posers and Puzzles 23 Jul '07 20:02
1. PBE6
Bananarama
23 Jul '07 20:021 edit
This one is borrowed from a website I came across this morning:

A card sharp walks up to you and says "I'll bet you 20:1 that you can't flip through this entire shuffled deck of cards while saying the card ranks in order and not have the rank you're saying match the rank of the card you just flipped over" (for example, if you flip "K, 7, 7, J, 5" while saying "A, 2, 3, 4, 5" you match the 5's on the 5th flip and you lose the bet).

Is this a fair offer? If not, what should the odds be?
2. uzless
The So Fist
23 Jul '07 20:072 edits
Originally posted by PBE6
This one is borrowed from a website I came across this morning:

A card sharp walks up to you and says "I'll bet you 20:1 that you can't flip through this entire shuffled deck of cards while saying the card ranks in order and not have the rank you're saying match the rank of the card you just flipped over" (for example, if you flip "K, 7, 7, J, 5" while sayin flip and you lose the bet).

Is this a fair offer? If not, what should the odds be?
no

25:1

*calc's

1 in 13 chance of naming the rank =1/13

multiply by 52 chances (cards)

=4% chance you will not say the correct rank

Which means you want at least 25:1 odds to break even

EDIT: this may change if you can see the cards as they are turned up so you have more information
3. PBE6
Bananarama
23 Jul '07 20:22
Originally posted by uzless
no

25:1

*calc's

1 in 13 chance of naming the rank =1/13

multiply by 52 chances (cards)

=4% chance you will not say the correct rank

Which means you want at least 25:1 odds to break even

EDIT: this may change if you can see the cards as they are turned up so you have more information
You're right, it's not a fair bet, but the odds should be higher than 25:1 to be fair...
4. DeepThought
24 Jul '07 14:10
Originally posted by uzless
no

25:1

*calc's

1 in 13 chance of naming the rank =1/13

multiply by 52 chances (cards)

=4% chance you will not say the correct rank

Which means you want at least 25:1 odds to break even

EDIT: this may change if you can see the cards as they are turned up so you have more information
I'm not convinced by this calculation, the problem is that the probability of the 5th card matching is not independent of the first card matching. The cards are in some unknown order. On the first card you have a 2 (say) and the odds of a match with the ace (playing ace low) is 12 to 1. This means that the odds of a 2 on the next card are 16 to 1, there are 51 cards out so there are 48 ways of not getting a match and 3 ways of getting a match. Having said that my intuition is that it's not a great bet.
5. 24 Jul '07 16:04
It's a card sharp anyway, he will have fixed it.
6. uzless
The So Fist
24 Jul '07 16:15
Originally posted by PBE6
You're right, it's not a fair bet, but the odds should be higher than 25:1 to be fair...
give us the explain
7. PBE6
Bananarama
24 Jul '07 17:47
Originally posted by uzless
give us the explain

8. uzless
The So Fist
24 Jul '07 18:00
Originally posted by PBE6

ah,

I multiplied by 52 instead of exponenting it 52 times.

9. wolfgang59
25 Jul '07 07:37
are wrong.

True odds of not making match are
(48/52)power13 * (47/51)power13 * (46/50)power13 * (45/49)power13

since after not making match with first ace there are now only 47 'non-aces' to select from 51 cards (et cetera). This is easy to think about if you consider calling the cards ace, ace, ace, ace, 2, 2, 2, 2, 3, 3, ..
10. Palynka
Upward Spiral
25 Jul '07 08:142 edits
Originally posted by wolfgang59
are wrong.

True odds of not making match are
(48/52)power13 * (47/51)power13 * (46/50)power13 * (45/49)power13

since after not making match with first ace there are now only 47 'non-aces' to select from 51 cards (et cetera). This is easy to think about if you consider calling the cards ace, ace, ace, ace, 2, 2, 2, 2, 3, 3, ..
I think you're on to something here, but I'm not convinced about the odds.

The AAAA22223333... ranking is an excellent way to think about it. My question is, what happens when you reach 2?

Before reaching 2:
(48/52) * (47/51) * (46/50) * (45/49)

Here we agree, right?

Reaching 2:
Here we need to factor these things
Ai - Prob an amount i of 2 were drawn in the first four, given that the first four were not an ace); i = 0,1,2,3,4
B - Prob of not drawing a 2 in the 2222 rank.

So we need to calculate
Prob(B|A0)*P(A0) + Prob(B|A1)*P(A1)+Prob(B|A2)*P(A2) + Prob(B|A3)*P(A3) + Prob(B|A4)*P(A4)

And similarly until the end...

I don't think each one of these are equal to:
(48/52) * (47/51) * (46/50) * (45/49) (which would eventually give us the power13)

What do you think?

PS: In fact, I think they cannot be. When we reach the last 4 cards (KKKK), the number of Kings in them is pre-determined by the 48 previous draws....

Edit - I also think that the odds at that link are wrong.

Edit 2, forgot that i can be zero A0.
11. wolfgang59
25 Jul '07 08:48
I think you are over complicating.

We agree on odds for not picking an ace.

We agree that how we pick is irrelevant (ie 1,2,3,4, .. is same as 1111,2222, 33)

Therefore odds of not picking any number correctly are the same for each and every number.

You are attempting to workout the odds of not drawing a 2 when a 2 was not drawn in first four cards and adding that to odds of not drawing a 2 when a 2 was drawn in first four cards.

Since the odds of drawing a 2 in first four cards added to odds of not drawing a 2 in first four cards is 1 we have no need to calculate those situations individually.
12. Palynka
Upward Spiral
25 Jul '07 09:01
Originally posted by wolfgang59
Since the odds of drawing a 2 in first four cards added to odds of not drawing a 2 in first four cards is 1 we have no need to calculate those situations individually.
We do because those odds are not neutral to the probability of drawing a 2 in the 2222 rank.

P(A0)+P(A1)+P(A2)+P(A3)+P(A4)+P(A5)=1

BUT

Prob(B|A0)*P(A0) + Prob(B|A1)*P(A1)+Prob(B|A2)*P(A2) + Prob(B|A3)*P(A3) + Prob(B|A4)*P(A4)
is different than
Prob(B|A0) + Prob(B|A1) +Prob(B|A2) + Prob(B|A3)*P(A3) + Prob(B|A4)

So the weights - P(Ai) - are important.
13. wolfgang59
25 Jul '07 11:43
?
The odds of a deuce coming up in 5th, 6th, 7th or 8th turns MUST be the same as an ace coming up in 1st, 2nd 3rd or 4th!!!! Forget the Math and look at the logic!

If I specify ANY turns of the cards (eg 5th, 9th, 20th, 51st) the odds of not getting an ace (OR ANY OTHER CARD) in those turns are the same; namely that which we previously agreed:-

(48/52) * (47/51) * (46/50) * (45/49)

It does not matter what showed in previous turns.
14. Palynka
Upward Spiral
25 Jul '07 12:022 edits
Originally posted by wolfgang59
?
The odds of a deuce coming up in 5th, 6th, 7th or 8th turns MUST be the same as an ace coming up in 1st, 2nd 3rd or 4th!!!! Forget the Math and look at the logic!

If I specify ANY turns of the cards (eg 5th, 9th, 20th, 51st) the odds of not getting an ace (OR ANY OTHER CARD) in those turns are the same; namely that which we previously agreed:-

(48/52) * (47/51) * (46/50) * (45/49)

It does not matter what showed in previous turns.
The odds of a deuce coming up in 5th, 6th, 7th or 8th turns MUST be the same as an ace coming up in 1st, 2nd 3rd or 4th!!!! Forget the Math and look at the logic!

I agree. The thing is that these events are not independent so the odds of both happening (or both not happening, in this case) is different than the simple product of the odds (again, of not happening).
15. 25 Jul '07 13:28
There's an additional complication to this. You can improve your chances by using an intelligent strategy.

For an extreme case, consider what happens if the first four cards are all Kings (and assume you haven't lost yet). You now know there are no more Kings. So if you keep saying "King" you win.